21 Mar 2019

# Visualize $e^{i\theta}$

$e^{i\theta}$ is the points on unit circle, at angle $\theta$, here is a visualization of how such circle is derived from the sum of power series:

Assume we have a function with the property of its derivative is equal to itself, just as $\frac{d}{dx}e^x = e^x$, such function should have the form:

Expand to complex number $i\theta$:

Draw these complex numbers as vectors on complex plane, we have a end point for each $\theta$, and a unit circle with all $\theta$s

move mouse horizontally to draw the circle

# Complex number as a Power Series

Any point on C plane can be written as

Let’s first compute the value of z using complex logarithm:

each term in the series is a vector, we add up these vectors to recover a in complex plane.

move mouse to draw the vectors

# Euler’s formula

$e^{iy} = \cos y + i \sin y$ corresponds to imagenary axis

# Compound interest

$e^{z} = e^{x+iy} = e^x e^{iy} = (1 + \frac{z}{n})^n$ if n is large

# Inverse

The inverse can be seen the image of z wrt. the unit circle, the mapping has some interesting trajectories, draw some strokes below to explore…

move mouse to draw 1/z
function drawETheta(theta, iter, scale) {
var itheta = new Complex(0, theta);
var res = new Complex(1, 0).add(itheta);

var path = new Path();
path.strokeColor = 'black';

var itheta_pow = new Complex(itheta);

for (var i = 2; i < iter; i++) {
itheta_pow = itheta_pow.multiply(itheta);
}
return path;
}

function drawExpZ(z, iter) {
// res = 1 + z
var w = z.log();
var res = new Complex(1 * scale, 0).add(w * scale);
var w_pow = new Complex(w);

var path = new Path();
path.strokeColor = 'black';